Question

If the lines $2x-3y=5$ and $3x-4y=7$ are two diameters of a circle of radius $7$, then the equation of the circle is ?

• A. $x^2+y^2+2x-4y-47=0$
• B. $x^2+y^2=49$
• C. $x^2+y^2-2x+2y-47=0$
• D. $x^2+y^2=17$

Answer 1

The correct option is C $x^2+y^2-2x+2y-47=0$

Let $(x_0,y_0)$ be the center of the circle

Intersection of any 2 diameter gives us the center.

$x=\frac{5+3y}{2}$ substituting this in

$3x–4y=7$

$3\frac{5+3y}{2}–4y=7$

$⟹y+1=0⟹y=-1$

$x_0=\frac{5+3(-1)}{2}=1$

Equation of circle will be

$(x–x_0{)}^2+(y–y_0{)}^2=r^2$

$(x–1{)}^2+(y+1{)}^2=7^2$

$⟹x^2+y^2-2x+2y–47=0$